Let I=_a^b(x^4-2x^2),dx.

Question

Let $I=\displaystyle\int_a^b(x^4-2x^2)\,dx$. If $I$ is minimum then the ordered pair $(a,b)$ is:

Accepted Answer

For $I$ to be minimum (as a function of endpoints), the integrand $x^4-2x^2$ should be $\le0$ throughout $[a,b]$, i.e. we minimize by choosing $[a,b]$ over the zeros of $x^4-2x^2=x^2(x^2-2)=0\Rightarrow x=0,\pm\sqrt2$.

Since $x^4-2x^2<0$ for $x\in(-\sqrt2,\sqrt2)\setminus\{0\}$, evaluating $I=\left[\dfrac{x^5}5-\dfrac{2x^3}3\right]_a^b$ at the candidate endpoint pairs shows the most negative (minimum) value occurs for $(a,b)=(-\sqrt2,\sqrt2)$, giving $I=-\dfrac{16\sqrt2}{15}$.

Solutions · 1